Question

Math

Posted 3 months ago

```
Let $g$ be a function defined for all real numbers except for $x=2$.
Also let $g^{\prime}$, the derivative of $g$, be defined as $g^{\prime}(x)=\frac{x^{2}}{(x-2)^{3}}$.
On which intervals is $g$ increasing?
Choose 1 answer:
(A) $(-\infty, 0)$ and $(0,2)$
(B) $(-\infty, 0)$ and $(2, \infty)$
(C) $(2, \infty)$ only
(D) $(0,2)$ only
(E) The entire domain of $g$
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 1

Determine the intervals where $g'(x) = \frac{x^2}{(x - 2)^3}$ is greater than zero

step 2

From the asksia-ll calculation list, the solution to $g'(x) > 0$ is $x > 2$

step 3

Since $g'(x) > 0$ for $x > 2$, the function $g$ is increasing on the interval $(2, \infty)$

Answer

(C) $(2, \infty)$ only

Key Concept

Intervals of Increase

Explanation

A function is increasing on intervals where its derivative is positive. The asksia-ll calculator determined that $g'(x) > 0$ for $x > 2$, so $g$ is increasing on the interval $(2, \infty)$.

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